cf Educational Codeforces Round 65 D. Bicolored RBS

原題：
D. Bicolored RBS
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

A string is called bracket sequence if it does not contain any characters other than “(” and “)”. A bracket sequence is called regular (shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters “+” and “1” into this sequence. For example, “”, “(())” and “()()” are RBS and “)(” and “(()” are not.

We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the 2
-nd pair lies inside the 1-st one, the 3-rd one — inside the 2-nd one and so on. For example, nesting depth of “” is 0, “()()()” is 1 and “()((())())” is 3.

Now, you are given RBS s of even length n. You should color each bracket of s into one of two colors: red or blue. Bracket sequence r, consisting only of red brackets, should be RBS, and bracket sequence, consisting only of blue brackets b, should be RBS. Any of them can be empty. You are not allowed to reorder characters in s, r or b

. No brackets can be left uncolored.

Among all possible variants you should choose one that minimizes maximum of r 's and b’s nesting depth. If there are multiple solutions you can print any of them.

Input

The first line contains an even integer n (2≤n≤2⋅10^5) — the length of RBS s.

The second line contains regular bracket sequence s (|s|=n, si∈{"(", “)”}).
Output

Print single string t of length n consisting of “0”-s and “1”-s. If ti is equal to 0 then character si belongs to RBS r, otherwise si belongs to b

.
Examples
Input

2
()

Output

11

Input

4
(())

Output
Copy

0101

Input
Copy

10
((()())())

Output
Copy

0110001111

Note

In the first example one of optimal solutions is s=
“()”. r is empty and b= “()”. The answer is max(0,1)=1

.

In the second example it’s optimal to make s=
“(())”. r=b= “()” and the answer is 1

.

In the third example we can make s=
“((()())())”. r= “()()” and b= “(()())” and the answer is 2.

中文：

給你一堆括號，然你讓把這個括號染成紅藍兩種顏色，定義括號的深度爲一個有效括號序列嵌套的最多層數。現在問你如何將括號染色，使得兩種顏色括號中深度最大值最小。

代碼：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
 
typedef pair<int,int> pii;
const int maxn = 200001;
 
string s;
int n;
int d[maxn];
stack<char> sc;
int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n)
    {
        cin>>s;
        int dep=0;
        memset(d,0,sizeof(d));
        for(int i=0;i<n;i++)
        {
            if(s[i]=='(')
            {
                sc.push('(');
                d[i]=sc.size();
                dep=max(dep,d[i]);
            }
            else
            {
                d[i]=sc.size();
                sc.pop();
            }
        }
        for(int i=0;i<n;i++)
        {
            if(d[i]<=dep/2)
                cout<<0;
            else
                cout<<1;
        }
        cout<<endl;
    }
    return 0;
}
 
 

思路：

大水題
問你怎麼把一組括號序列分成兩組，而且兩組序列深度的值最小。
先計算處原始括號序列的深度dep，那麼要想使得分成兩組後，兩組括號序列深度最大的最小，就要使兩個括號的深度儘量相同。 所以，兩組括號序列深度的值就是dep/2